Question: $f(x) = x^{2}+6x$ $g(n) = -4n^{2}-3(h(n))$ $h(n) = -2n^{2}-4n+2(f(n))$ $ g(h(-1)) = {?} $
Solution: First, let's solve for the value of the inner function, $h(-1)$ . Then we'll know what to plug into the outer function. $h(-1) = -2(-1)^{2}+(-4)(-1)+2(f(-1))$ To solve for the value of $h$ , we need to solve for the value of $f(-1)$ $f(-1) = (-1)^{2}+(6)(-1)$ $f(-1) = -5$ That means $h(-1) = -2(-1)^{2}+(-4)(-1)+(2)(-5)$ $h(-1) = -8$ Now we know that $h(-1) = -8$ . Let's solve for $g(h(-1))$ , which is $g(-8)$ $g(-8) = -4(-8)^{2}-3(h(-8))$ To solve for the value of $g$ , we need to solve for the value of $h(-8)$ $h(-8) = -2(-8)^{2}+(-4)(-8)+2(f(-8))$ To solve for the value of $h$ , we need to solve for the value of $f(-8)$ $f(-8) = (-8)^{2}+(6)(-8)$ $f(-8) = 16$ That means $h(-8) = -2(-8)^{2}+(-4)(-8)+(2)(16)$ $h(-8) = -64$ That means $g(-8) = -4(-8)^{2}+(-3)(-64)$ $g(-8) = -64$